Difference between revisions of "Conjugate Root Theorem"

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=Theorem=
 
=Theorem=
 
The Conjugate Root Theorem states that if <math>P(x)</math> is a polynomial with real coefficients, and <math>a+bi</math> is a root of the equation <math>P(x) = 0</math>, where <math>i = \sqrt{-1}</math>, then <math>a-bi</math> is also a root.
 
The Conjugate Root Theorem states that if <math>P(x)</math> is a polynomial with real coefficients, and <math>a+bi</math> is a root of the equation <math>P(x) = 0</math>, where <math>i = \sqrt{-1}</math>, then <math>a-bi</math> is also a root.
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==Uses==
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This has many uses. If you get a fourth degree polynomial, and you are given that a number in the form of <math>a+bi</math> is a root, then you know that <math>a-bi</math> in the root. Using the [[Factor Theorem]], you know that <math>(x-(a+bi))(x-(a-bi))</math> is also a root. Thus, you can multiply that out, and divide it by the original polynomial, to get a depressed quadratic equation. Of course, it doesn't have to be a fourth degree polynomial. It could just simplify it a bit.
  
 
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Revision as of 23:14, 10 January 2012

Theorem

The Conjugate Root Theorem states that if $P(x)$ is a polynomial with real coefficients, and $a+bi$ is a root of the equation $P(x) = 0$, where $i = \sqrt{-1}$, then $a-bi$ is also a root.

Uses

This has many uses. If you get a fourth degree polynomial, and you are given that a number in the form of $a+bi$ is a root, then you know that $a-bi$ in the root. Using the Factor Theorem, you know that $(x-(a+bi))(x-(a-bi))$ is also a root. Thus, you can multiply that out, and divide it by the original polynomial, to get a depressed quadratic equation. Of course, it doesn't have to be a fourth degree polynomial. It could just simplify it a bit.

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